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- To show there is no point of \(\mathcal{E}(\mathbb{Q}_p)\) reducing to the singular point \((0,0) \in \widetilde{\mathcal{E}}_p(\mathbb{F}_p)\), consider the reduction map:
\[
\mathrm{red}: \mathcal{E}(\mathbb{Q}_p) \to \widetilde{\mathcal{E}}_p(\mathbb{F}_p).
\]
A point \((x, y) \in \mathcal{E}(\mathbb{Q}_p)\) reduces to \((0,0)\) if and only if \(|x|_p < 1\) and \(|y|_p < 1\). Suppose such a point exists. Substituting into the equation \(y^2 = x^3 + p\), we analyze valuations:
- Since \(|x|_p < 1\), \(|x^3|_p = |x|_p^3 < 1\). Thus, \(|x^3 + p|_p = \max(|x^3|_p, |p|_p) = |p|_p = p^{-1}\) because \(|x^3|_p < |p|_p\).
- The left-hand side \(|y^2|_p = |y|_p^2\) must equal \(p^{-1}\), implying \(|y|_p = p^{-1/2}\). However, valuations in \(\mathbb{Q}_p\) are integers, and \(p^{-1/2}\) is not an integer. This contradiction shows no such \((x, y)\) exists.
- Deduce that $\# \mathcal{E}(\mathbb{Q})_{\text {tors }}=1$.
The exact sequence for the reduction map at a prime \(p\) of bad reduction is:
\[
0 \rightarrow \mathcal{E}_1(\mathbb{Q}_p) \rightarrow \mathrm{red}^{-1}\left(\widetilde{\mathcal{E}}_p(\mathbb{F}_p)^{\mathrm{ns}}\right) \rightarrow \widetilde{\mathcal{E}}_p(\mathbb{F}_p)^{\mathrm{ns}} \rightarrow 0.
\]
From part 1, \(\mathcal{E}(\mathbb{Q}_p) = \mathrm{red}^{-1}\left(\widetilde{\mathcal{E}}_p(\mathbb{F}_p)^{\mathrm{ns}}\right)\). The subgroup \(\mathcal{E}_1(\mathbb{Q}_p)\) is torsion-free, so \(\mathcal{E}(\mathbb{Q})_{\text{tors}}\) injects into \(\widetilde{\mathcal{E}}_p(\mathbb{F}_p)^{\mathrm{ns}}\), which has order \(p\).
\(\#\mathcal{E}(\mathbb{Q})_{\text{tors}}\) also divides \(\#\widetilde E_5(\mathbb{F}_5)=6\). Since \(p\) and \(6\) are coprime, \(\mathcal{E}(\mathbb{Q})_{\text{tors}}\) must be trivial.
gr.inc/question/c-i-show-that-there-is-no-poi … eftmathbbq_pright-re |
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